Question

If sin theta =6/10 then the value of tan theta +sec theta

Answer 1

Given:-

• $\sf \sin{ \theta} = \dfrac{6}{10}$

To find:-

• $\sf Value\;of\; \tan{ \theta} + \sec{ \theta}$

Solution:-

We have, $\sf \sin{ \theta} = \dfrac{6}{10} = \dfrac{3}{5}$

$\sf \sin{ \theta} = \dfrac{3}{5} = \dfrac{Height}{Hypotenuse}$

$\star\sf {\underline{Using\;Pythagoras\;Theorem:-}}$

$\dashrightarrow\sf H^2 = B^2 + P^2$

$\dashrightarrow\sf (5)^2 = B^2 + (3)^2$

$\dastrightarrow\sf 25 = B^2 + 9$

$\dashrightarrow\sf B^2 = 25 - 9$

$\dashrightarrow\sf B^2 = 16$

$\dashrightarrow\sf \sqrt{B^2} = \sqrt{16}$

$\dashrightarrow\;{\sf{\purple{B = 4}}}$

$\rule{200}3$

Therefore,

$\sf \tan{ \theta} = \dfrac{height}{base} = \dfrac{4}{3}$

And

$\sf \sec{ \theta} = \dfrac{hypotenuse}{base} = \dfrac{5}{4}$

Now,

Add the values of $\sf \tan{ \theta} \; and \; \sec{ \theta} :-$

$\dashrightarrow\sf \tan{ \theta} + \sec{ \theta}$

$\dashrightarrow\sf \dfrac{4}{3} + \dfrac{5}{4}$

$\dashrightarrow\sf \dfrac{16 + 15}{12}$

$\dashrightarrow\;{\sf{\red{\dfrac{31}{12}}}}$

$\dag\;\sf Hence\;Solved!!$

$\rule{200}3$

Answer 2

Answer:

Required value is 2.

Step-by-step explanation:

Given $\sin(\Theta) = \frac{6}{10}$

Now we want to find $\tan(\Theta) + \sec(\Theta)$

First here we will find relation between $\tan(\Theta)$and $\sin(\Theta)$

and second we will find relation between$\sin(\Theta) \:$and $\sec(\Theta)$

Now,

$\tan(\Theta) = \frac{ \sin(\Theta) }{ \cos(\Theta) } = \frac{ \sin(\Theta) }{ \sqrt{1 - { \sin(\Theta) }^{2} } } = \frac{ \frac{6}{10} }{ \sqrt{1 - {( \frac{6}{10} )}^{2} } } = \frac{ \frac{6}{10} }{ \sqrt{1 - \frac{36}{100} } } = \frac{ \frac{6}{10} }{ \sqrt{ \frac{64}{100} } } = \frac{ \frac{6}{10} }{{ \frac{8}{10} } } = \frac{6}{10} \times \frac{10}{8} = \frac{3}{4}$

And

$\sec(\Theta) = \frac{1}{ \cos(\Theta) } = \frac{1}{ \sqrt{1 - { \sin(\Theta) }^{2} } } = \frac{1}{ \sqrt{1 - {( \frac{6}{10}) }^{2} } } = \frac{1}{ \sqrt{ \frac{64}{ 100} } } = \frac{10}{8} = \frac{5}{4}$

So,

$\tan(\Theta) + \sec(\Theta) = \frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2$

Here applied formulas are

${sin}^{2} x + {cos}^{2} x = 1$

$\tan(x) = \frac{ \sin(x) }{ \cos(x) }$

Required value is 2.